# Induction on n

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1. Jan 19, 2016

### bubblewrap

Below I have uploaded the page I am having trouble with.

Here it says that it is using induction on n but I don't understand how it uses the formula for when n=j to derive the formula for when n=j+1

2. Jan 19, 2016

### Samy_A

What is the definition of $y_n(t)$?
I can guess that equation (5) is $|y_n(t)-y_0| \leq M(t-t_0)$. Is this guess correct?

EDIT: ok, I found it, $y_n$ are Picard iterates.
Equation (5) is $|y_n(t)-y_0| \leq M(t-t_0)$ for $t_0 \leq t \leq t_0+\alpha$.
So by definition $y_{j+1}(t)=y_0 + \int_{t_0}^t f(s,y_j(s)) \, ds$

He uses the definition of $y_{j+1}$ and that $M$ is defined as the maximum on the rectangle defined by $t_0 \leq t \leq t_0+a,\ |y-y_0| \leq b$.
Thanks to the induction hypothesis that he knows that $|f(s,y_j(s))|\leq M$.

Last edited: Jan 19, 2016
3. Jan 19, 2016

### bubblewrap

But he didn't use n=j to show anything since $f(s,y_j(s))\leq M$ is a definition and not something deduced from n=j

4. Jan 19, 2016

### Samy_A

Let's rewrite it in the correct order.
By definition, $y_{j+1}(t)=y_0 + \int_{t_0}^t f(s,y_j(s)) \, ds$.
Let R be the rectangle defined by $t_0 \leq t \leq t_0+a,\ |y-y_0| \leq b$ where $a, b$ are positive numbers.
Let $M$ be the maximum of $f$ on that rectangle.
Set $\alpha=min(a,\frac{b}{M})$.
Then $|y_n(t)-y_0| \leq M(t-t_0)$ for $t_0 \leq t \leq t_0+\alpha$. (Equation (5) in your text.)

Assume that it has been proven for $j$.
So $|y_j(t)-y_0|\leq M(t-t_0)$ for $t_0 \leq t \leq t_0+\alpha$.
But then $|y_j(t)-y_0|\leq M(t-t_0) \leq M\alpha \leq b$, meaning that $(t,y_j(t))$ lies in the rectangle R, so that $|f(t,y_j(t)| \leq M$. This is where the induction hypothesis is crucial.
That's what he uses in the last step: in the integral he just uses $|f(s,y_j(s)|\leq M$.

This only holds if $(s,y_j(s))$ lies in the rectangle R. He knows it does thanks to the induction hypothesis.

Last edited: Jan 19, 2016
5. Jan 19, 2016

### bubblewrap

Thank you for your help, one last thing, am I slow if I am not getting this?

6. Jan 19, 2016

### Samy_A

Do you agree that we are only certain that $|f(s,y_j(s)| \leq M$ if $(s,y_j(s))$ lies in the rectangle R?

7. Jan 19, 2016

### bubblewrap

Rectangle R depends on the two constants a and b, and these are random numbers right? So although we are not certain about what you said it could be in that condition.

8. Jan 19, 2016

### Samy_A

Yes, a and b are two positive constants. They define the rectangle R. M is by definition the maximum of f on that rectangle.

To use that $|f(s,y_j(s))| \leq M$, you must know that $(s,y_j(s))$ lies in the rectangle R. Do you agree so far?

9. Jan 19, 2016

### bubblewrap

Yes, sorry I just had dinner. I agree so far.

10. Jan 19, 2016

### Samy_A

No reason to be sorry for having dinner.

Assume that equation (5) has been proved for j.
Now he needs to show that $(s,y_j(s))$ lies in the rectangle R if $t_0 \leq s \leq t_0+\alpha$.
Equation (5) means: $|y_j(s)-y_0| \leq M(s-t_0)$ for $t_0 \leq s \leq t_0+\alpha$.
But then $|y_j(s)-y_0| \leq M(s-t_0) \leq M\alpha \leq b$, by the definition of $\alpha$. (*)

Remember how the rectangle R was defined?
$(s,y) \in R$ if $t_0 \leq s \leq t_0+a,\ |y-y_0| \leq b$.

Now look at $(s,y_j(s))$ for $t_0 \leq s \leq t_0+\alpha$.
Obviously $t_0 \leq s \leq t_0+a$.
But, more interesting, by (*), $|y_j(s)-y_0| \leq b$.
We conclude that $(s,y_j(s)) \in R$, and then by definition of M, $|f((s,y_j(s))| \leq M$. This is what he uses in the last step of his proof. He needed the induction hypothesis to prove that $(s,y_j(s)) \in R$.

11. Jan 19, 2016

### bubblewrap

Thank you it's all clear now :) Is the book not self explanatory or am I slow for not getting it in the first place? (I got it now)

12. Jan 19, 2016

### Samy_A

It's the book. This point is not very difficult, but one line explaining this would have been welcome.
Now he does explain just before the proof that the lemma means that the graph of $y_n$ stays within the rectangle: